1. if ab ≡ −1 (mod 24), then a + b ≡ 0 (mod 24).

2. If n ≡ −1 (mod 24), then it is not a perfect square, and the number of divisors is even.

3. If n ≡ −1 (mod 24), then 24 divides $\displaystyle \sum_{d|n}d$

I don't need complete proofs, just some hints would be enough.